Integrand size = 43, antiderivative size = 213 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {(9 A-14 B+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 \sqrt {a} d}+\frac {\sqrt {2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {(7 A-2 B+8 C) \tan (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-6 B) \sec (c+d x) \tan (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}} \]
-1/8*(9*A-14*B+8*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d/a ^(1/2)+(A-B+C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/ 2))*2^(1/2)/d/a^(1/2)+1/8*(7*A-2*B+8*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2 )-1/12*(A-6*B)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/3*A*sec(d* x+c)^2*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)
Time = 1.34 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.77 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (48 (A-B+C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3(c+d x)-3 \sqrt {2} (9 A-14 B+8 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3(c+d x)+(37 A-6 B+24 C-4 (A-6 B) \cos (c+d x)+3 (7 A-2 B+8 C) \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{24 d \sqrt {a (1+\cos (c+d x))}} \]
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[a + a*Cos[c + d*x]],x]
(Cos[(c + d*x)/2]*Sec[c + d*x]^3*(48*(A - B + C)*ArcTanh[Sin[(c + d*x)/2]] *Cos[c + d*x]^3 - 3*Sqrt[2]*(9*A - 14*B + 8*C)*ArcTanh[Sqrt[2]*Sin[(c + d* x)/2]]*Cos[c + d*x]^3 + (37*A - 6*B + 24*C - 4*(A - 6*B)*Cos[c + d*x] + 3* (7*A - 2*B + 8*C)*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(24*d*Sqrt[a*(1 + C os[c + d*x])])
Time = 1.48 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.11, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.372, Rules used = {3042, 3522, 27, 3042, 3463, 27, 3042, 3463, 27, 3042, 3464, 3042, 3128, 219, 3252, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a \cos (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {\int -\frac {(a (A-6 B)-a (5 A+6 C) \cos (c+d x)) \sec ^3(c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {(a (A-6 B)-a (5 A+6 C) \cos (c+d x)) \sec ^3(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a (A-6 B)-a (5 A+6 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{6 a}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {\int -\frac {3 \left (a^2 (7 A-2 B+8 C)-a^2 (A-6 B) \cos (c+d x)\right ) \sec ^2(c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a}+\frac {a (A-6 B) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}}{6 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-6 B) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \int \frac {\left (a^2 (7 A-2 B+8 C)-a^2 (A-6 B) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-6 B) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \int \frac {a^2 (7 A-2 B+8 C)-a^2 (A-6 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-6 B) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {\int -\frac {\left (a^3 (9 A-14 B+8 C)-a^3 (7 A-2 B+8 C) \cos (c+d x)\right ) \sec (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {a^2 (7 A-2 B+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-6 B) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (a^3 (9 A-14 B+8 C)-a^3 (7 A-2 B+8 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-6 B) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (9 A-14 B+8 C)-a^3 (7 A-2 B+8 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3464 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-6 B) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (9 A-14 B+8 C) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-16 a^3 (A-B+C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-6 B) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (9 A-14 B+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-16 a^3 (A-B+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-6 B) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (9 A-14 B+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {32 a^3 (A-B+C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-6 B) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (9 A-14 B+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {16 \sqrt {2} a^{5/2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-6 B) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {-\frac {2 a^3 (9 A-14 B+8 C) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {16 \sqrt {2} a^{5/2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-6 B) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^{5/2} (9 A-14 B+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {16 \sqrt {2} a^{5/2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}\) |
(A*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) - ((a*(A - 6*B)*Sec[c + d*x]*Tan[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) - (3*(-1/2* ((2*a^(5/2)*(9*A - 14*B + 8*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*C os[c + d*x]]])/d - (16*Sqrt[2]*a^(5/2)*(A - B + C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d)/a + (a^2*(7*A - 2*B + 8*C) *Tan[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]])))/(4*a))/(6*a)
3.5.8.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A *b - a*B)/(b*c - a*d) Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Simp[(B*c - A*d)/(b*c - a*d) Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m* (c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C - B*d)*( a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2* (n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ [m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(2040\) vs. \(2(184)=368\).
Time = 15.44 (sec) , antiderivative size = 2041, normalized size of antiderivative = 9.58
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(2041\) |
| default | \(\text {Expression too large to display}\) | \(2398\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+cos(d*x+c)*a)^(1/2),x, method=_RETURNVERBOSE)
1/6*A*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*a*(16*2^(1/2) *ln(4*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))-9*ln( 4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a* sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))-9*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^ (1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2 )*a^(1/2)-2*a)))*sin(1/2*d*x+1/2*c)^6+(576*2^(1/2)*ln(4*(a^(1/2)*(a*sin(1/ 2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*a-324*ln(4/(2*cos(1/2*d*x+1/2 *c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2 )^(1/2)*a^(1/2)+2*a))*a-324*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)* a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))* a+168*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))*sin(1/2*d*x+1/2*c)^4 +(-288*2^(1/2)*ln(4*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x +1/2*c))*a+162*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+ 1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a+162*ln(-4/(2 *cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin( 1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a-160*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^ 2)^(1/2)*a^(1/2))*sin(1/2*d*x+1/2*c)^2+48*2^(1/2)*ln(4*(a^(1/2)*(a*sin(1/2 *d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*a-27*ln(4/(2*cos(1/2*d*x+1/2*c )+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^ (1/2)*a^(1/2)+2*a))*a-27*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*...
Time = 0.39 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.47 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {3 \, {\left ({\left (9 \, A - 14 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (9 \, A - 14 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, {\left (7 \, A - 2 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (A - 6 \, B\right )} \cos \left (d x + c\right ) + 8 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) + \frac {48 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{4} + {\left (A - B + C\right )} a \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{96 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(1 /2),x, algorithm="fricas")
1/96*(3*((9*A - 14*B + 8*C)*cos(d*x + c)^4 + (9*A - 14*B + 8*C)*cos(d*x + c)^3)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*sqrt(a*cos(d* x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(3*(7*A - 2*B + 8*C)*cos(d*x + c)^2 - 2*(A - 6*B)* cos(d*x + c) + 8*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c) + 48*sqrt(2)*((A - B + C)*a*cos(d*x + c)^4 + (A - B + C)*a*cos(d*x + c)^3)*log(-(cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(a) - 2*cos(d* x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3)
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Timed out} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(1 /2),x, algorithm="maxima")
Time = 0.45 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.70 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\frac {24 \, \sqrt {2} {\left (A \sqrt {a} - B \sqrt {a} + C \sqrt {a}\right )} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {24 \, \sqrt {2} {\left (A \sqrt {a} - B \sqrt {a} + C \sqrt {a}\right )} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {3 \, {\left (9 \, A - 14 \, B + 8 \, C\right )} \log \left ({\left | \frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {3 \, {\left (9 \, A - 14 \, B + 8 \, C\right )} \log \left ({\left | -\frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} {\left (84 \, A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 24 \, B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 96 \, C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 96 \, C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{48 \, d} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(1 /2),x, algorithm="giac")
1/48*(24*sqrt(2)*(A*sqrt(a) - B*sqrt(a) + C*sqrt(a))*log(sin(1/2*d*x + 1/2 *c) + 1)/(a*sgn(cos(1/2*d*x + 1/2*c))) - 24*sqrt(2)*(A*sqrt(a) - B*sqrt(a) + C*sqrt(a))*log(-sin(1/2*d*x + 1/2*c) + 1)/(a*sgn(cos(1/2*d*x + 1/2*c))) - 3*(9*A - 14*B + 8*C)*log(abs(1/2*sqrt(2) + sin(1/2*d*x + 1/2*c)))/(sqrt (a)*sgn(cos(1/2*d*x + 1/2*c))) + 3*(9*A - 14*B + 8*C)*log(abs(-1/2*sqrt(2) + sin(1/2*d*x + 1/2*c)))/(sqrt(a)*sgn(cos(1/2*d*x + 1/2*c))) - 2*sqrt(2)* (84*A*sqrt(a)*sin(1/2*d*x + 1/2*c)^5 - 24*B*sqrt(a)*sin(1/2*d*x + 1/2*c)^5 + 96*C*sqrt(a)*sin(1/2*d*x + 1/2*c)^5 - 80*A*sqrt(a)*sin(1/2*d*x + 1/2*c) ^3 - 96*C*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 + 27*A*sqrt(a)*sin(1/2*d*x + 1/2* c) + 6*B*sqrt(a)*sin(1/2*d*x + 1/2*c) + 24*C*sqrt(a)*sin(1/2*d*x + 1/2*c)) /((2*sin(1/2*d*x + 1/2*c)^2 - 1)^3*a*sgn(cos(1/2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^4\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \]